A bicycle lock requires a two-digit code of numbers 1 through 9, and any digit may be used only once. Which expression would determine the probability that both digits are even?
Answers
The expression which would determine the probability that both digits are even which is required for bicycle lock is (4P1)(3P1)/(9P2).
What do you understand by the term permutation?
The permutation is the arrangement of the things or object in a systematic order, in all the possible ways. The order of arrangement in permutation is in linear.
A bicycle lock requires a two-digit code of numbers 1 through 9, and any digit may be used only once. The probability of choosing 2 digits from 9 is,
There are total 4 even numbers {2,4,6,8}. The probability of choosing first digit's even from 4 even numbers is,
For the second digit to be even is,
Thus, the favorable outcome is, and total outcome is
. Thus, the expression which would determine the probability that both digits are even is,
Thus, the expression which would determine the probability that both digits are even which is required for bicycle lock is (4P1)(3P1)/(9P2).
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The correct answer is:
A. P(both even) =
The expression would determine the probability that both digits are even.
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What is 3-3 times 6+2=
Ellason wants to solve the following system using the elimination method: 5x + 2y = 30 x + y = 8 What number should the equation x + y = 8 be multiplied by to eliminate y?
$70 interested at 12% compunded continuously after a period of 4 years
Answers
Y^2 = R^2 - X^2
Y = sqrt(R^2 - X^2)
Answer:
Y = √(R² - X²)
Step-by-step explanation:
A is equal to B complement.
A∪B={}
A is not a subset of B.
Answers
If A∩B = Ø and A and B are not the empty set, all of the following are true except Option 3
What is an empty set?
That set which doesn't contain any value is called empty set. This is also called as null set or void set. This is denoted by Ø or { }
We have given A∩B = Ø
Case1:
If A∩B = Ø and A and B are not the empty set, A and B are non empty set which means A and B are disjoint.
Hence, Option 1 is correct.
Case2:
If there intersection is empty means there is nothing common. Hence, Option 2 is correct.
Case3:
A and B are non-empty sets means their union can not be empty.
Hence, option 3 is incorrect.
Case4:
A can't be subset of B because Subset is the same elements in B of A which is not possible because their intersection is empty.
Hence, Option 4 is correct.
Therefore, If A∩B = Ø and A and B are not the empty set, all of the following are true except Option 3.
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Answer:
Option 3
Step-by-step explanation:
We have give
Case1:
A and B are non empty set which means A and B are disjoint
Hence, Option 1 is correct.
Case2:
When there intersection is empty means there is nothing common
Hence, A is not of B or A is equal to complement of B
Hence, Option 2 is correct.
Case3:
A and B are non-empty sets means their union can not be empty
Hence, option 3 is incorrect.
Case4:
A can not be subset of B because Subset is the same elements in B of A which is not possible because their intersection is empty
Hence, Option 4 is correct.
We have to choose false statement
Therefore, Option 3
PLEASE HELP!!!!!!!!
Answers
b) (y-1)^2 = 20(x-3)
c) (y-1)^2 = -20(x-3)
d) (x-3)^2 = 20(y-1)
Answers
Answer with Step-by-step explanation:
We have to find:
the standard form of the equation of the parabola that has a vertex of (3, 1) and a directrix of x = –2
General form of Parabola that opens left or right:
(y−k)²=4p(x−h)
Vertex =(h,k)
Directrix: x=h−p
Here, h=3,k=1 and h-p=-2 i.e. p=h+2=5
Hence, equation of parabola in this case equals
(y-1)²=4×5(x-3)
i.e. (y-1)²=20(x-3)
Hence, correct option is:
b) (y-1)²=20(x-3)
Remenber:
if S=(0,0) y²=2px then F=(p/2,0 and d:x=-p/2
Let's pose
x'=x-3
y'=y-1 axes passing by the point (3,0) in base (x,y) and (0,0) in base (x',y')
y'²=2p'x'
d': x'=-5=-p'/2 ==> p'=10 and y²=20x'
Returning in base (x,y) : (y-1)²=20(x-3)
Answer B
With step by step explanation !
Answers
It's easier to deal with the symbolic sum (in sigma notation),
Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives
and it's the latter two terms that reveal a telescoping pattern.
In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that
or
which gives a = 1, and a + b = 0 so that b = -1.
Our sum has been rearranged as
The remaining telescoping sum is
1/2 [(1/1 - 1/2) + (1/2- 1/3) + (1/3- 1/4) + … + (1/48- 1/49) + (1/49- 1/50) + (1/50 - 1/51)]
and you can see how there are pairs of numbers that cancel, so that the sum reduces to
1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51
So, our original sum ends up being
Answers
a-44=16
a=60
Sarah is 60 years old.
Hope this helps.
Answer:
60
4(a-44)=64
Step-by-step explanation: